10 exam-style questions on equations reducible by substitution (homogeneous, exact, missing variable, etc.). Tap Show answer for full working.
Solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{y}{x}\) by the substitution \(y=vx\).
\(y=vx\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}=v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}\). So \(v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}=v\Rightarrow x\dfrac{\mathrm{d}v}{\mathrm{d}x}=0\Rightarrow v=C\). Thus \(y=Cx\).
Use the substitution \(v=y'\) to solve \(y''=2y'\), given \(y(0)=1\), \(y'(0)=2\).
\(v'=2v\Rightarrow\dfrac{\mathrm{d}v}{\mathrm{d}x}=2v\Rightarrow\dfrac{\mathrm{d}v}{v}=2\,\mathrm{d}x\Rightarrow\ln|v|=2x+C\Rightarrow v=Ae^{2x}\).
\(y'=Ae^{2x}\). \(y'(0)=2\Rightarrow A=2\). \(y=\mathrm{e}^{2x}+B\). \(y(0)=1\Rightarrow B=0\). \(y=\mathrm{e}^{2x}\).
Show that the substitution \(v=\dfrac{y}{x}\) reduces the equation \(x\dfrac{\mathrm{d}y}{\mathrm{d}x}=y+x\) to a separable equation, and solve it.
\(y=vx\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}\). Left: \(x(v+x v')=vx+x\Rightarrow x^2v'=x\Rightarrow xv'=1\Rightarrow\dfrac{\mathrm{d}v}{\mathrm{d}x}=\dfrac{1}{x}\). \(v=\ln|x|+C\). \(y=x\ln|x|+Cx\).
Solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{x+y}{x-y}\) using a homogeneous substitution.
Let \(y=vx\). Then \(v+x v'=\dfrac{1+v}{1-v}\). \(x v'=\dfrac{1+v}{1-v}-v=\dfrac{1+v-v+v^2}{1-v}=\dfrac{1+v^2}{1-v}\).
\(\dfrac{1-v}{1+v^2}\,\mathrm{d}v=\dfrac{\mathrm{d}x}{x}\). Integrate: \(\arctan v-\dfrac{1}{2}\ln(1+v^2)=\ln|x|+C\). Back-substitute \(v=y/x\).
The differential equation \(x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=y^2+xy+x^2\) is homogeneous. Solve it by the substitution \(y=vx\).
\(v+x v'=\dfrac{v^2+v+1}{1}\). \(x v'=v^2+1\). \(\dfrac{\mathrm{d}v}{v^2+1}=\dfrac{\mathrm{d}x}{x}\). \(\arctan v=\ln|x|+C\). \(y=x\tan(\ln|x|+C)\).
Solve \(y''+y'=0\) by reducing the order, given \(y(0)=0\), \(y'(0)=1\).
Let \(v=y'\). \(v'+v=0\Rightarrow v=Ae^{-x}\). \(y' =Ae^{-x}\). \(y=-Ae^{-x}+B\). \(y'(0)=1\Rightarrow A=1\). \(y(0)=0\Rightarrow -1+B=0\Rightarrow B=1\). \(y=1-\mathrm{e}^{-x}\).
Use the substitution \(v=x+y\) to solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=(x+y)^2\).
\(v=x+y\Rightarrow\dfrac{\mathrm{d}v}{\mathrm{d}x}=1+\dfrac{\mathrm{d}y}{\mathrm{d}x}=1+v^2\). \(\dfrac{\mathrm{d}v}{1+v^2}=\mathrm{d}x\). \(\arctan v=x+C\). \(v=\tan(x+C)\). \(y=\tan(x+C)-x\).
Solve the equation \(x\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=xy^2\) by a suitable substitution (Bernoulli or otherwise).
Divide by \(x\): \(\dfrac{\mathrm{d}y}{\mathrm{d}x}+\dfrac{1}{x}y=y^2\). Bernoulli: let \(z=y^{-1}\). Then \(\dfrac{\mathrm{d}z}{\mathrm{d}x}=-\dfrac{1}{y^2}\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Multiply original by \(-y^{-2}\): \(-\dfrac{1}{y^2}y'-\dfrac{1}{x}y^{-1}=-1\). So \(\dfrac{\mathrm{d}z}{\mathrm{d}x}-\dfrac{1}{x}z=-1\).
Integrating factor \(\dfrac{1}{x}\). \(\dfrac{1}{x}z=\int-\dfrac{1}{x}\,\mathrm{d}x=-\ln|x|+C\). \(z=-x\ln|x|+Cx\). \(y=\dfrac{1}{x(C-\ln|x|)}\).
The equation \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+\dfrac{1}{x}\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\) has a solution independent of \(x\) in the first derivative. Reduce the order and solve fully.
Let \(v=y'\). Then \(v'+\dfrac{1}{x}v=0\Rightarrow\dfrac{\mathrm{d}v}{v}=-\dfrac{\mathrm{d}x}{x}\Rightarrow\ln|v|=-\ln|x|+C\Rightarrow v=\dfrac{A}{x}\).
\(y'=A/x\Rightarrow y=A\ln|x|+B\).
Solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2x+y+1}{x+2y-1}\) by a translation of axes to remove the constant terms.
Solve the simultaneous system \(2x+y+1=0\), \(x+2y-1=0\) to find the intersection point of the lines. Solving: \(x= -1\), \(y=1\). Shift: let \(X=x+1\), \(Y=y-1\).
Then \(\dfrac{\mathrm{d}Y}{\mathrm{d}X}=\dfrac{2X+Y}{X+2Y}\). This is homogeneous. Substitute \(Y=vX\) and solve as before, then translate back.